Question

Let ABC be a triangle with sides 51, 52, 53. Let

$\alpha$


denote the incircle of ∆ABC. Draw tangents to

$\alpha$


which are parallel to the sides of ABC. Let r1,r2,r3 be the inradii of the three corner triangles so

formed. Find the largest integer that does not exceed r1 + r2 + r3.​

Answer 1

Step-by-step explanation:

In geometry, the incircle or inscribed circle of a triangle is the largest circle contained in the triangle; it touches (is tangent to) the three sides. The center of the incircle is a triangle center called the triangle's incenter.[1]

Answer 2

The largest integer that does not exceed r1 + r2 + r3 is 50.

Given:

Triangle ABC with sides 51, 52, 53 respectively

To Find:

Largest integer that don't exceed r1 + r2 + r3

Solution:

We have the sides of the triangle ABC as 51, 52, and 53.

The semi-perimeter of triangle ABC, denoted as s, is given by:

s = (51 + 52 + 53) / 2 = 78.

Using Heron's formula, the area of the triangle ABC is given by:

A = √(s(s-a)(s-b)(s-c)) = √(782726*25) = 702√14.

The radius r of the incircle α is given by:

r = A / s = 702√14 / 78 = 9√14 / 2.

Let r1, r2, and r3 be the radii of the incircles of the triangles formed by the tangents to α which are parallel to the sides of ABC.

By considering the two tangents to the incircle from a vertex of the triangle, the radii r1, r2, and r3 are given by:

r1 = r2 + r3 - a,

r2 = r1 + r3 - b,

r3 = r1 + r2 - c,

where a, b, and c are the sides of triangle ABC.

Substituting the values of a, b, and c, we get:

r1 = r2 + r3 - 51,

r2 = r1 + r3 - 52,

r3 = r1 + r2 - 53.

Adding the above equations, we obtain:

r1 + r2 + r3 = (r1 + r2 + r3) - 50.

Hence,

r1 + r2 + r3 = 50.

Therefore, the largest integer that does not exceed r1 + r2 + r3 is 50.

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